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**Q11. ABC is a right triangle, right-angled at A, and D is the mid-point of AB. Prove that BC ^{2} = CD^{2} + 3 BD^{2}. **

**Sol. **We have a right Δ ABC in which <A = 90°

∴Using Pythagoras Theorem, we have:

BC^{2} = AB^{2} + AC^{2} ...(1)

Again, Δ ACD is right D, <A = 90°

∴CD^{2} = AD^{2} + AC^{2}

...(2) [Using Pythagoras Theorem]

Subtracting (2) from (1), we get

BC^{2} - CD^{2} = AB^{2} - AD^{2} ...(3)

Since D is the mid-point of AB

∴2 BD = AB and AD = BD ...(4)

From (3) and (4), we have:

∴BC^{2} - CD^{2} = (2 BD)^{2} - (BD)^{2}

= 4 BD^{2} - BD^{2}

BC^{2} = CD^{2} + 3 BD^{2}**Q12. In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC. **

**Sol.** Let us draw EOF y AB ⇒ OE ⊥ AD and OF ⊥ BC

In Δ OFB

Œ <F = 90°

∴Using Pythagoras theorem, we have:

CB^{2} = OF^{2} + BF^{2} ...(1)

In Δ OED

<E = 90°

∴Using Pythagoras theorem, we have:

OD^{2} = OE^{2} + DE^{2} ...(2)

Adding (1) and (2), we get

OB^{2} + OD^{2} = OF^{2} + BF^{2} + OE^{2} + DE^{2}

= OF^{2} + AE^{2} + OE^{2} + CF^{2}

[Œ BF = AE and CF = DE]

= (OF^{2} + CF^{2}) + (OE^{2} + AE^{2})

= OC^{2} + OA^{2}

= OC^{2} + 52

⇒ 62 + 82 = OC^{2} + 52

⇒ 36 + 64 = OC^{2} + 25

⇒ OC^{2} = 36 + 64 - 25 = 75

⇒

Thus OC = 5√3cm.**Q13. In the figure, if AD ⊥ Bc, then prove that: AB**

**Sol.** In D aDc, <ADc = 90°

∴ AD^{2} = AC^{2} - CD^{2} .....(1) (Using Pythagoras Theorem)

Similarly, in D AbD,

⇒ AD^{2} = AB^{2} - DB^{2}.....(2)

From (1) and (2), we have

AB^{2} - DB^{2} = AC^{2} - CD^{2}

⇒ AB^{2} + CD^{2} = AC^{2} + BD^{2}**Q14. In the given figure, AD ⊥ BC and BD = CD. Prove that:**

∴ 3 BD = CD

Since BD + DC = BC

∴ BD + 3 BD = BC

⇒ 4 BD = BC

⇒ BD = 1/4 BC

⇒ CD = 3/4 BC

Now, in right Δ ADC, < D = 90°

By Pythagoras theorem, we get

CA

Also in right Δ ADB

AD

From (1) and (2),

CA

**Sol. **We have parallelogram ABCD in which M is the mid point of CD.

In Δ EMD and Δ BMC

MD = MC [Œ M is mid-point of CD]

<EMD = <CMB

[Vertically opposite angles]

<MED = <MBC

[Alternate interior angles]

∴ Δ BMC ≌ Δ EMD [AAS congruency]

⇒ BC = ED ⇒ AD = ED ...(1)

[Œ BC = AD, opposite sides of parallelogram]

Now, in Δ AEL and Δ CBL

<AEL = <CBL [Alternate interior angles]

<ALE = <CLB

[Vertically opposite angles]

∴ By AA similarity, we have:**Q16. In the given figure, Δ ABC is right-angled at C and DE ^ AB. Prove that Δ ABC ~ Δ ADE and hence find the length of AE and DE.**

Sol.In Δ ABC and Δ ADE, we have:

<A = <A [Common]

<C = <E [Each = 90°]

∴ Δ ABC ~ Δ ADE [AA Similarity]

...(1)

In right Δ ABC, <C = 90°

Using Pythagoras theorem, we have:

AB

= 122 + 52

= 144 + 25 = 169

⇒AB = √169 = 13 cm

Now, from (1), we get

⇒ DE = = 2.77 cm

And =

⇒ AE

**Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE ^{2} = BD × EC.**

**Sol.** In Δ DBG and Δ ECF

<3 + <1 = 90° = <3 + <4

∴<3 + <1 = <3 + <4

⇒ <1 = <4

<D = E = 90°

∴Using AA similarity, we have:

=

BD × EC = EF × DG

But DG = EF = DE

∴BD × EC = DE × DE

⇒ BD × EC = DE^{2}

Thus, DE^{2} = BD × EC**Q18. In the figure, AD ⊥ BC and BD = CD. Prove that 2 CA ^{2} = 2 AB^{2} + BC^{2}. **

**Sol.** Œ BD = 1/3 CD

⇒ 3 BD = CD

∴BC = BD + DC

⇒ BC = BD + 3 BD

⇒ BC = 4 BD ...(1)

And

From (2) ...(4)

In right Δ ADC,

Using Pythagoras theorem,

CA^{2} = AD^{2} + DC^{2}

=

From (3) ...(4)

In right Δ ADB

Using Pythagoras theorem,

AD^{2} = AB^{2} - BD^{2}

= **Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.**

**Sol.** We have a quadrilateral ABCD such that its diagonals intersect at O and

=

<AOB = <COD [Vertically opposite angles]

∴ Using SAS similarity, we have

Δ AOB ~ Δ COD

⇒ Their corresponding angles are equal.

i.e., <1 = <2

But they form a pair of int. alt. angles.

⇒ AB y DC

⇒ ABCD is a trapezium.**Q20. Two triangles ABC and DBC are on the same base BC and on the same side of BC in which <A = <D = 90°. If CA and BD meet each other at E, show that AE· EC = BE· ED Sol.** We have right Δ ABC and right Δ DBC on the same base BC such that

<A = <D = 90°

In Δ ABE and Δ DCE

<A = <D = 90°

<1 = <2 [Vertically opp. angles]

∴ Using AA similarity, we have:

Δ ABE ~ Δ DCE

⇒ Their corresponding sides are proportional.

⇒

⇒ AE· EC = BE· ED**Q21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB Sol. **We have parallelogram ABCD

In Δ ABE and Δ CFB, we have

<A = <C

[Opposite angles of parallelogram]

<AEB = <EBC

[Alternate angles, AD y BC]

∴ Using AA similarity, we get

Δ ABE ~ Δ CFB

**Q22. In Δ ABC, if AD is the median, then show that AB ^{2} + AC^{2} = 2 [AD^{2} + BD^{2}]. **

∴ BD = DC

Let us draw AE ≌ BC

Now, in rt. Δs AEB and AEC, we have

AB

AC

Adding (1) and (2),

AB

= (BD - ED)2 + AE2 +

(CD + DE)2 + AE2

= 2AE

= 2 [AE

[BD = CD]

= 2 [AD]2 + 2BD

[AE

= 2 [AD

Thus, AB

AC

Sol.

∴ BC = 2 BD

Now, in Δ ABC

AC

= (2 BD)

= 4 BD

In right Δ ABD

Using Pythagoras theorem,

AD

⇒ BD

From (1) and (2), we get

AC

⇒ AC

⇒ AC

or AC

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