Question

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that $\angle DBC$ is a right angle.

Answer 1

Hint: First prove the congruence of $\Delta \text{AMC and }\Delta \text{DMB}$ then using this congruence condition and the properties of parallel lines with a transversal can prove the $\angle \text{DBC}$ as 90°.

Complete step-by-step answer:

Given that M is the mid-point of AB so BM = AM. It is also given that DM = CM.
DC and AB intersect each other so$\angle \text{DMB = }\angle \text{AMC}$ (vertically opposite angles).
Hence, ∆AMC ≅ ∆DMB (By SAS congruence)

Now, $\angle \text{ACM = }\angle \text{BDM}$ (By CPCT (Corresponding part of Congruent Triangles)). As we have shown that $\angle \text{ACM = }\angle \text{BDM}$ and AC = DB (By CPCT). As we can see that $\angle \text{ACM and }\angle \text{BDM }$ are alternate interior angles in which DC is transversal and AC is parallel to DB. Hence, we can say AC is parallel to DB.
Now, $\angle \text{DBC and }\angle \text{ACB}$are interior angles if BC is transversal and AC is parallel to DB.
We know that interior angles on the same side of a transversal are supplementary or sum of interior angles on the same side of the transversal is 180° and it is given that $\angle \text{ACB}$ is 90°.
$\begin{align}
  & \angle \text{DBC + }\angle \text{ACB = 18}{{\text{0}}^{0}} \\
 & \angle \text{DBC + 9}{{\text{0}}^{0}}={{180}^{0}} \\
 & \angle \text{DBC = 9}{{\text{0}}^{0}} \\
\end{align}$
Hence, we have shown that $\angle \text{DBC}$ is a right angle.

Note: We cannot assume from the figure that$\angle \text{ACM and }\angle \text{BDM}$ are alternate interior angles, we need to prove it which we have done using properties of congruent triangles.