Question

In the adjacent figure ABCD is a square and $\vartriangle APB$ is an equilateral triangle. Prove that $\vartriangle APD\tilde = \vartriangle BPC$

Answer 1

Hint:In $\vartriangle APD$ and $\vartriangle BPC$
 Use SAS rule (side angle side)
SAS rule states that for any 2 triangles if two sides and the included angle are equal then the triangles are said to be congruent.

Complete step-by-step answer:
Given, ABCD is a square
is an equilateral triangle
Then,
$\overline {AP} = \overline {BP} $ ( Sides of an equilateral triangle $\vartriangle APB$ )
$\overline {AD} = \overline {BC} $ (same side of square)
And $\angle DAP = \angle DAB - \angle PAB = {90^ \circ } - {60^ \circ } = {30^ \circ }$
$\angle DAP = {30^ \circ }$--- 1
$\therefore \angle DAP = \angle BPC$
$\therefore $$\vartriangle APD$is congruent to $\vartriangle BPC$ through SAS rule.
In $\vartriangle APD$
$AP = AD$ as$AP = AB$ (Equilateral triangle)
We know that $\angle DAP = {30^ \circ }$
$\therefore \angle APD = \dfrac{{\left( {{{180}^ \circ } - {{30}^ \circ }} \right)}}{2}$
($\vartriangle APD$ is an Isosceles triangle and $\angle APD$is one of the base angles)
$\angle APD = \dfrac{{{{150}^ \circ }}}{2} = {75^ \circ }$
Similarly, $\angle BPC = {75^ \circ }$
$\therefore $$
  \angle DPC = {360^ \circ } - \left( {{{75}^ \circ } + {{75}^ \circ } + {{60}^ \circ }} \right) \\
  \angle DPC = {150^ \circ } \\
 $
Now in $\vartriangle PDC$
$PC = PD$ as $\vartriangle APD$ is congruent to $\vartriangle BPC$
$\therefore \vartriangle PDC$is an isosceles triangle.
And $
  \angle PCD = \angle PDC = \dfrac{{{{180}^ \circ } + {{150}^ \circ }}}{2} = {15^ \circ } \\
  \angle DPC = {150^ \circ } \\
  \angle PDC = {15^ \circ } \\
  \angle PCD = {15^ \circ } \\
 $
Hence proved.

Note:In this type of questions students will face problems in finding the angle. So be careful while finding the angle. Students should focus on diagrams which shall help them in solving the questions or proving the given statement.