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**Which of the following pairs of triangles are congruent by ASA condition?**

1) We have

Since âˆ ABO = âˆ CDO = 45Â° and both are alternate angles,

ABâˆ¥DC

âˆ BAO = âˆ DCO (alternate angle , ABâˆ¥CD and AC is a transversal line)

âˆ ABO = âˆ CDO = 45Â° (given in the figure)

Also, AB = DC (Given in the figure)

Therefore, by ASA â–³AOB â‰…â–³DOC

2)

In â–³ABC ,

Now AB = AC (Given)

âˆ ABD = âˆ ACD = 40Â° (Angles opposite to equal sides)

âˆ ABD +âˆ ACD+âˆ BAC=180Â° (Angle sum property)

40Â°+40Â°+âˆ BAC=180Â°

âˆ BAC=180Â°-80Â°=100Â°

âˆ BAD +âˆ DAC=âˆ BAC

âˆ BAD=âˆ BAC-âˆ DAC=100Â°-50Â°=50Â°

âˆ BAD =âˆ CAD = 50Â°

Therefore, by ASA, â–³ABD â‰…â–³ADC

3)

InÎ”ABC,

âˆ A+âˆ B+âˆ C=180Â°(Angle sum property)

âˆ C=180Â°-âˆ A-âˆ B

âˆ C=180Â°-30Â°-90Â°=60Â°

InÎ”PQR,

âˆ P+âˆ Q+âˆ R=180Â°(Angle sum property)

âˆ P=180Â°-âˆ Q-âˆ R

âˆ P=180Â°-60Â°-90Â°=30Â°

âˆ BAC = âˆ QPR = 30Â°

âˆ BCA=âˆ PRQ = 60Â°

and AC = PR (Given)

Therefore, by ASA, â–³ABC â‰…â–³PQR

4)

We have only BC =QR but none of the angles of â–³ABC AND â–³PQR are equal.

Therefore, â–³ABCâ‰† â–³PRQ

**In Fig. 37, AD bisects âˆ A and AD âŠ¥ BC.**

(i) Is âˆ† ADB â‰… âˆ† ADC?

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

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

(i) Yes, â–³ADB â‰…â–³ADC, by ASA criterion of congruency

(ii) We have used âˆ BAD =âˆ CAD

âˆ ADB=âˆ ADC = 90Â° since ADâŠ¥BC

and AD = DA

(iii) Yes, BD= DC since, â–³ADB â‰…â–³ADC

**Draw any triangle ABC. Use ASA condition to construct another triangle congruent to it.**

We have drawn

â–³ABC with âˆ ABC = 60Â°and âˆ ACB = 70Â°

We now construct â–³PQRâ‰…â–³ABC

â–³PQR has âˆ PQR =60Â° and âˆ PRQ = 70Â°

Also we construct â–³PQR such that BC =QR

Therefore by ASA the two triangles are congruent

**In âˆ† ABC, it is known that âˆ B = âˆ C. Imagine you have another copy of âˆ† ABC**

(i) Is âˆ† ABC â‰… âˆ† ACB?

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

(iii) Is it true to say that AB = AC?

(i) Yesâ–³ABC â‰…â–³ACB

(ii) We have used âˆ ABC=âˆ ACB and âˆ ACB =âˆ ABC again

Also BC = CB

(iii) Yes, it is true to say that AB = AC since âˆ ABC=âˆ ACB

**In Fig. 38, AX bisects âˆ BAC as well as âˆ BDC. State the three facts needed to ensure that âˆ† ABD â‰… âˆ† ACD.**

As per the given conditions,âˆ CAD=âˆ BAD and âˆ CDA=âˆ BDA (because AX bisects âˆ BAC ) AD=DA (common)

Therefore, by ASA, â–³ACDâ‰…â–³ABD

**Question 6:**

In Fig. 39, *AO* = *OB* and âˆ *A* = âˆ *B*.

**(i) Is âˆ† AOC â‰… âˆ† BOD?**

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that âˆ ACO = âˆ BDO?

**Answer 6:**

We have

âˆ OAC =âˆ OBD, AO = OBAlso, âˆ AOC = âˆ BOD (Opposite angles on same vertex)

Therefore, by ASA â–³AOC â‰…â–³BOD