Question

In fig AP and DP are the bisectors of two adjacent angles A and D of quadrilateral ABCD. prove that 2∠APD=∠B+∠C.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Final Answer:

For AP and DP to be the bisectors of two adjacent angles $\angle A$ and $\angle D$ of quadrilateral ABCD, it is proved that $2\angle APD=\angle B+\angle C$, as shown in the attached figure.

Given:

AP and DP are the bisectors of two adjacent angles $\angle A$ and $\angle D$ of quadrilateral ABCD.

To Find:

In the quadrilateral ABCD, it is to be proved that $2\angle APD=\angle B+\angle C$.

Explanation:

The following points are significant to arrive at the solution to the present problem.

• The sum of the four interior angles of any quadrilateral is equal to three hundred and sixty degrees.
• The angle bisector of any angle, as the name suggests, bisects the referred angle into two equal parts.

Step 1 of 4

As the statement in the given problem suggests, note the following.

As AP and DP are the bisectors of two adjacent angles $\angle A$ and $\angle D$ of quadrilateral ABCD,

• The bisector AP of the angle $\angle A$ and $\angle D$ of quadrilateral ABCD, divides it into two equal angles; such that$\angle BAP=\angle PAD=\frac{\angle BAD}{2}$.
• The bisector DP of the angle $\angle D$ of quadrilateral ABCD, divides it into two equal angles; such that $\angle CDP=\angle PDA=\frac{\angle CDA}{2}$.

Step 2 of 4

In accordance with the above calculations, the following is derived.

In the triangle $\triangle APD$,

$\angle PAD+\angle PDA+\angle APD=180\textdegree\\\angle APD=180\textdegree-(\angle PAD+\angle PDA)\\\angle APD=180\textdegree-(\frac{\angle BAD}{2}+\frac{\angle CDA}{2})$

Step 3 of 4

Now, in the quadrilateral ABCD,

$\angle ABC+\angle BCD+\angle CDA+\angle BAD=360\textdegree\\\angle CDA+\angle BAD=360\textdegree-(\angle ABC+\angle BCD)$

Step 4 of 4

Finally, the above two derivations are combined to get the following.

$\angle APD\\=180\textdegree-(\frac{\angle BAD}{2}+\frac{\angle CDA}{2})\\=180\textdegree-(\frac{\angle BAD}{2}+\frac{\angle CDA}{2})\\=180\textdegree-(\frac{\angle BAD+\angle CDA}{2})\\=180\textdegree-\frac{360\textdegree-(\angle ABC+\angle BCD)}{2}\\=\frac{(\angle ABC+\angle BCD)}{2}$

Hence, it is proved that  $2\angle APD=(\angle ABC+\angle BCD)$.

Therefore, it is proved that $2\angle APD=\angle B+\angle C$, as shown in the attached figure, where AP and DP are the bisectors of two adjacent angles $\angle A$ and $\angle D$ of quadrilateral ABCD.

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