Question

In this given figure, ABCD and BPQ line. BP = BC and DQ || CP. Prove that
(I) CP = CD
(II) DP bisects angle CDQ.

Answer 1

Hint: In this question, we are given BP = BC and DQ || CP. Also, $\mathrm{\angle }ABP=4x\text{ and }\mathrm{\angle }CPD=x$ We need to prove that CP = CD and DP bisects $\mathrm{\angle }CDQ$ which means $\mathrm{\angle }CDP=\mathrm{\angle }PDQ$. For this, we will use the following properties of the triangle, lines, and angles:
(I) Exterior angle product: Exterior angle of a triangle is equal to the sum of the interior opposite angles.
(II) Isosceles triangle property: Angles corresponding to equal sides is an isosceles triangle and are always equal.
(III) Converse of isosceles triangle property: If two angles are equal in a triangle, then the corresponding side of equal angles are also equal.
(IV) If two lines are parallel and a transversal forms corresponding angles with parallel lines then corresponding angles are equal.

Complete step-by-step solution
Here we are given the diagram as:

Here BP = BC, $$DQ\parallel CP,\mathrm{\angle }ABP=4x\text{ and }\mathrm{\angle }CPD=x$$.
(i) Firstly, we need to prove CP = CD.
As we can see from triangle BPC that BP = BC, therefore $\mathrm{\angle }BPC=\mathrm{\angle }BCP$ because of isosceles triangle property, according to which angles corresponding to equal sides are also equal. Hence,
$\mathrm{\angle }BPC=\mathrm{\angle }BCP\cdots \cdots \cdots \left(1\right)$.
Now as we can see $\mathrm{\angle }ABP$ acts as an exterior angle to $\mathrm{\Delta }BPC$.
So using exterior angle property, which states that, an exterior angle is equal to the sum of two opposite interior angles, we get:
$\mathrm{\angle }ABP=\mathrm{\angle }BPC+\mathrm{\angle }BCP$.
From (1), $\mathrm{\angle }BPC=\mathrm{\angle }BCP$.
Also, $\mathrm{\angle }ABP=4x$ so we get:
$4x=\mathrm{\angle }BCP+\mathrm{\angle }BCP\Rightarrow 4x=2\mathrm{\angle }BCP\Rightarrow 2x=\mathrm{\angle }BCP$
Hence $\mathrm{\angle }BCP=\mathrm{\angle }BPC=2x$.
Now using exterior angle property in the $\mathrm{\Delta }PCD$ where $\mathrm{\angle }BCP$ is the exterior angle and $\mathrm{\angle }CPD\text{ and }\mathrm{\angle }CDP$ are opposite interior angles, we get $\mathrm{\angle }BCP=\mathrm{\angle }CPD+\mathrm{\angle }CDP$.
As we know, $\mathrm{\angle }BCP=2x\text{ and }\mathrm{\angle }CPD=x$ so we get:
$2x=x+\mathrm{\angle }CPD\Rightarrow \mathrm{\angle }CPD=2x-x\Rightarrow \mathrm{\angle }CPD=x$.
Hence in the triangle $\mathrm{\Delta }CPD$, $\mathrm{\angle }CPD=\mathrm{\angle }CDP=x$.
As we know, from the converse of the isosceles triangle property, sides corresponding to equal angles are also equal. Therefore, CP = CD. Hence proved.
(ii) Now let us prove DP bisects $\mathrm{\angle }CDQ$.
Hence we need to prove $\mathrm{\angle }CDP=\mathrm{\angle }PDQ$.
From the diagram, we know that CP || DQ.
Since CP and DQ are parallel lines and CD acts as transversal, so $\mathrm{\angle }PCB\text{ and }\mathrm{\angle }QDC$ will form a pair of corresponding angles. As we know, corresponding angles are always equal, therefore $\mathrm{\angle }PCB=\mathrm{\angle }QDC$.
As equivalent earlier, $\mathrm{\angle }PCB=2x$ so, $\mathrm{\angle }QDC=2x$.
Now, as we can see, $\mathrm{\angle }PDC+\mathrm{\angle }QDP=\mathrm{\angle }QDC$ so $\mathrm{\angle }PDC+\mathrm{\angle }QDP=2x$.
Also we know $\mathrm{\angle }PDC=x$ so,
$x+\mathrm{\angle }QDP=2x\Rightarrow \mathrm{\angle }QDP=2x-x\Rightarrow \mathrm{\angle }QDP=x$.
Now $\mathrm{\angle }QDP=\mathrm{\angle }CDP$ therefore DP bisects $\mathrm{\angle }CDQ$. Hence proved.

Note: Students should know all the properties of triangles before solving these sums. While using exterior angle property, make sure to take the sum of opposite interior angles only. We can take $\mathrm{\angle }QDP=\mathrm{\angle }CPD$ by considering PD as transversal for parallel lines PC and DQ where $\mathrm{\angle }QPD\text{ and }\mathrm{\angle }CPD$ are alternate interior angles.