Question

In a $\Delta ABC$, the internal bisector of $\angle A$ meets opposite side BC at D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Which of the following is/are true?

• A. $\Delta ACE$ is isosceles.
• B. $AD=\frac{1}{2}EC$
• C. $\frac{AB}{BD}=\frac{AC}{DC}$
• D. $\angle ACE+\angle AEC+\angle ABC={180}^{\circ }-\angle ACD$

Answer 1

Answer: (A), (C), (D)

The correct options are
A

$\Delta ACE$ is isosceles.

C

$\frac{AB}{BD}=\frac{AC}{DC}$

D

$\angle ACE+\angle AEC+\angle ABC={180}^{\circ }-\angle ACD$

Given: $AD\parallel EC$,
$\angle BAD=\angle DAC$

$\angle BAD=\angle AEC$ [Corresponding angles] ..... (1)

$\angle DAC=\angle ACE$ [Alternate angles] ..... (2)

And we know that $\angle BAD=\angle DAC$,

So, $\angle AEC=\angle ACE$

$\Rightarrow \Delta AEC$ is isosceles triangle.

Using intercept theorem,

$\frac{AB}{AE}=\frac{BD}{DC}[AD\parallel EC\text{in}\Delta BCE]$

$\frac{AB}{AC}=\frac{BD}{DC}$ [AE = AC]

$\Rightarrow \frac{AB}{BD}=\frac{AC}{DC}$

Since, A and D aren't mid points, we can't say

$AD=\frac{1}{2}EC$

In $\Delta ABC$,

$\angle ABD+\angle BAD+\angle DAC+\angle ACD={180}^{\circ }$

Using (1) and (2),

$\angle ABD+\angle ACE+\angle AEC={180}^{\circ }-\angle ACD$.