Question

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$
. Show that

(i) Δ DAP $\cong$ Δ EBP

(ii) AD = BE

Answer 1

(i)

In Δ DAP and Δ EBP, we have
$$\begin{gathered} \angle EPA=\angle DPB \\ \Rightarrow \angle EPA+\angle EPD=\angle DPB+\angle EPD\left(\mathrm{Adding}\angle EPD\mathrm{on}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow \angle APD=\angle EPB \end{gathered}$$
$$\begin{gathered} \angle DAP=\angle EBP\left(\mathrm{Given}\right) \\ AP=BP\left(\mathrm{Given}\right) \\ \therefore ∆DAP\cong ∆EBP\left(\mathrm{BY}\mathrm{ASA}\right) \end{gathered}$$
Hence, $∆DAP\cong ∆EBP$.

(ii)

$$\begin{gathered} \because ∆DAP\cong ∆EBP \\ \therefore AD=BE\left(\mathrm{By}\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}\right) \end{gathered}$$