Q32 of 34 Page 252

In the given figure, ABC is a triangle in which AB = AC. D is a point in the interior of ΔABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC.

In ΔBDC ∠DBC = ∠DCB so

BD = DC …(i)


{sides opposite to equal angles in a triangle are equal}


Now let’s consider that ΔABD and ΔADC –


AB = AC {given}


AD is a common side.


And BD = DC {from equation (i)}


Hence ΔABD and ΔADC are congruent.


So ∠BAD = ∠DAC (congruency criteria)


Hence AD bisects ∠BAC.


More from this chapter

All 34 →
30

On a graph paper plot the following points:

A(3, 3), B(2, 4), C(5, 5), D(0, 2), E(3, -3) and F(-5, -5).


Which of these points are the mirror images in (i) x-axis (ii) y-axis?

 31

In the given figure, in a ΔABC, BE ⊥ AC, ∠EBC = 40° and ∠DAC = 30. ∠DAC = 30°. Find the values of x, y and z.

 33

In the given figure, ABCD is a square and EF is parallel to diagonal DB and EM = FM. Prove that: (i) BF = DE (ii) AM bisects ∠BAD.

 34

In the given figure, AB || CD If ∠BAE = 100° and ∠ECD = 120° then x = ?