Question

ABC and DBC are two isosceles triangles on the same side of BC. Then DA (or AD ) produced bisects the BC at the right angle.
If the statement is true mention 1 otherwise 0.

Answer 1

Hint: Prove the congruence of the pair of triangles DBA and DCA and then DBE and DCE to conclude BE=CE where E is the point where DA intersects BC. Prove the congruence of triangles ABE and ACE to prove $\mathrm{\angle }DEB=\mathrm{\angle }DEC$and use the sum of angles on a straight line to check whether they are right angles.

Complete step-by-step solution:

We have drawn the figure as required by the question where the triangles ABC and DBC are on the same side of BC. It is given in the question that the triangles ABC and DBC are isosceles where the equal sides in $\mathrm{\Delta }\text{ABC}$ equal sides are AB=AC and equal sides in $\mathrm{\Delta }\text{DBC}$ are DB=DC. DA produced cuts BC and we named the point of intersection as E.$$$$
We are asked to check whether $\mathrm{\angle }DEB,\mathrm{\angle }DEC$ are right angles $\left({90}^{\circ }\right)$ or and BE=CE not. $$$$
We observe the triangles DBA and DCA where we already have AB=AC and DB=DC. They also have common side AD. So by side-side-side congruence we get $\mathrm{\Delta }\text{DBA}\cong \mathrm{\Delta }\text{DCA}$. So, the respective angles in both the triangles will be equal. So $\mathrm{\angle }ADB=\mathrm{\angle }ADC$. $$$$
We again observe the triangles DBE and DCE where we already have DB=DC ,$\mathrm{\angle }ADB=\mathrm{\angle }ADC\Rightarrow \mathrm{\angle }EDB=\mathrm{\angle }EDC$(as E and A are on the same line)and the common side DE. We use Side-Angle-Side congruence and have $\mathrm{\Delta }\text{DBE}\cong \mathrm{\Delta }\text{DCE}$. Now, the respective sides in both the triangles will be equal. So BE=CE$$$$
We observe the triangles ABE and CDE where we already have BE=CE(proved above), AB=AC and the common side AE. Using side-side-side congruence we have $\mathrm{\Delta }\text{ABE }\cong \mathrm{\Delta }\text{CDE}$. We equate the respective angles to get $\mathrm{\angle }AEB=\mathrm{\angle }AEC\Rightarrow \mathrm{\angle }DEB=\mathrm{\angle }DEC$(as A and D are on the same line). The sum of the angles $\mathrm{\angle }DEB,\mathrm{\angle }DEC$ is a straight angle$\left({180}^{\circ }\right)$ as the B, E and C lie on the same line. So we have,$$$$
$$\begin{array}{rl}& \mathrm{\angle }DEB+\mathrm{\angle }DEC={180}^{\circ }\\ & \Rightarrow \mathrm{\angle }DEC+\mathrm{\angle }DEC={180}^{\circ }(\because \mathrm{\angle }DEB=\mathrm{\angle }DEC)\\ & \Rightarrow \mathrm{\angle }DEC={\displaystyle \frac{{180}^{\circ }}{2}}={90}^{\circ }=\mathrm{\angle }DEB\end{array}$$
We have $\mathrm{\angle }DEB=\mathrm{\angle }DEC={90}^{\circ }$ and BE=CE. Hence it is true that DA bisects the side BC at right angle. So the answer is 1.

Note: The important thing to be careful of here is different types of congruence. If we pick any combination from sides and angles of one triangle and find them to be equal to sides or angles of another triangle they are not necessarily congruent, for example, angle-angle-angle equality where the triangles are similar but not congruent.