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**In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State each result in symbolic form. (Fig. 46)**

i)

âˆ ADC= âˆ BCA =90Â°

AD = BC

and hyp AB = hyp AB

Therefore, by RHS, â–³ADBâ‰…â–³ACB.

.

ii)

AD=AD (Common)

hyp AC = hyp AB (Given)

âˆ ADB+âˆ ADC = 180Â°(Linear pair)

âˆ ADB+90Â°=180Â°

âˆ ADB=180Â°âˆ’90Â°=90Â°

âˆ ADB=âˆ ADC = 90Â°

Therefore, by RHS, â–³ADB=â–³ADC

iii)

hyp AO = hyp DO

BO = CO

âˆ B =âˆ C =90Â°

Therefore, by RHS, â–³AOBâ‰…â–³DOC

iv)

Hyp AC = Hyp CA

BC= DC

âˆ ABC=âˆ ADC = 90Â°

Therefore, by RHS, â–³ABCâ‰…â–³ADC

v)

BD = DB

Hyp AB =Hyp BC, as per the given figure.

âˆ BDA +âˆ BDC = 180Â°

âˆ BDA +90Â° = 180Â°

âˆ BDA= 180Â°âˆ’90Â°= 90Â°

âˆ BDA =âˆ BDC = 90Â°

Therefore, by RHS, â–³ABDâ‰…â–³CBD

**âˆ† ABC is isosceles with AB = AC. AD is the altitude from A on BC.**

(i) Is âˆ† ABD â‰… ACD?

(ii) State the pairs of matching parts you have used to answer (i).

(ii) Is it true to say that BD = DC?

(i)Yes, â–³ABDâ‰…â–³ACDâ–³ABDâ‰…â–³ACD by RHS congruence condition.

(ii) We have used Hyp AB = Hyp AC

AD = DA

and âˆ ADB = âˆ ADC = 90Â°âˆ ADB = âˆ ADC = 90Â° (ADâŠ¥BC at point D)

(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.

**âˆ† ABC is isoseles with AB = AC. Also, AD âŠ¥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condtion do you use? Which side of âˆ† ADC equls BD? Which angle of âˆ† ADC equals âˆ B?**

We have AB = AC ......(1)

AD = DA (common)........(2)

and âˆ ADC=âˆ ADBâˆ ADC=âˆ ADB (ADâŠ¥BC at point D)........(3)

Therefore from 1, 2 and 3, by RHS congruence condition,

â–³ABDâ‰…â–³ACD

Now, the triangles are congruent .

Therefore, BD= CD.

And âˆ ABD=âˆ ACD (c.p.c.t).

**Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.**

Consider

â–³ABC with âˆ B as right angle.

We now construct another right triangle on base BC, such that âˆ C is a right angle and AB = DC

Also, BC = CB

Therefore, by RHS, â–³ABCâ‰…â–³DCB

**In Fig. 47, BD and CE are altitudes of **

(i) Is âˆ†

(ii) State the three pairs of matching parts you have used to answer (i).

(i) Yes, â–³BCDâ‰…â–³CBE by RHS congruence condition.

(ii) We have used hyp BC = hyp CB

BD = CE (given in question)

and âˆ BDC = âˆ CEB =90Â°