#### Exercise 16.4

**Q1. Which of the following pairs of triangle are congruent by ASA condition?**

**Answer: **

i)

We have,

Since \(\angle\)

\(\angle\)

Therefore, by ASA \(\Delta AOB \cong \Delta DOC\)

ii)

In ABC ,

Now AB =AC (Given)

\(\angle\)

\(\angle\)

40Â° + 40Â° + \(\angle\)

\(\angle\)

\(\angle\)

\(\angle\)

Therefore, by ASA, \(\Delta ABD \cong \Delta ADC\)

iii)

In \(\Delta\)

\(\angle\)

\(\angle\)

In PQR,

\(\angle\)

\(\angle\)

\(\angle\)

\(\angle\)

Therefore, by ASA, \(\Delta ABC \cong \Delta PQR\)

iv)

We have only

BC = QR but none of the angles of \(\Delta\)

Therefore, \(\Delta ABC Â and cong \Delta PRQ\)

**Q2. In figure, AD bisects A and AD and AD \(\perp\) BC.**

**(i) Is \(\Delta ADB \cong \Delta ADC\)?**

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

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

Answer:

(i)Â Yes, \(\Delta ADB \cong \Delta ADC\)

(ii) We have used \(\angle\)

Since, AD \(\perp\)

(iii) Yes, BD = DC since, \(\Delta ADB \cong \Delta ADC\)

**Q3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.**

Answer:

We have drawn

\(\Delta\)

We now construct \(\Delta PQR \cong \Delta ABC\)

Also we construct \(\Delta\)

Therefore by ASA the two triangles are congruent

**Q4. In \(\Delta\) ABC, it is known that \(\angle\)Â B = C. Imagine you have another copy ofÂ \(\Delta\) ABCÂ **

**(i) Is \(\Delta ABC \cong \Delta ACB\)**

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

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

**Answer: **

(i) Yes \(\Delta ABC \cong \Delta ACB\)

(ii) We have used \(\angle\)

Also BC = CB

(iii) Yes it is true to say that AB = AC since \(\angle\)

**Q5. In Figure, AX bisects \(\angle\) BAC as well as \(\angle\)Â BDC. State the three facts needed to ensure that \(\Delta ACD \cong \Delta ABD\)**

**Answer: **

As per the given conditions, \(\angle\)

AD = DA (common)

Therefore, by ASA, \(\Delta ACD \cong \Delta ABD\)

**Q6. In Figure, AO = OB and \(\angle\) A = \(\angle\) B.Â **

**(i) Is \(\Delta AOC \cong \Delta BOD\)**

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

**(iii) Is it true to say that \(\angle\) ACO = \(\angle\)BDO?**

**Answer:**

We have

\(\angle\)

AO = OB

Also, \(\angle\)

Therefore, by ASA \(\Delta AOC \cong \Delta BOD\)