Congruent Triangles Geometry Maths Tuition: Solution

congruent-tuition-2

Solution:

(a) \angle EBD=\angle BEC (given)

BE is a common side for both triangles \triangle BCE and \triangle EDB

BD=EC (given)

Therefore, \triangle BCE \equiv \triangle EDB (SAS)

(proved)

(b)

Since \triangle BCE\equiv\triangle EDB we have \angle CBE=\triangle DEB

Thus, \begin{array}{rcl}\angle ABE&=&180^\circ - \angle CBE\\    &=&180^\circ -\angle DEB\\    &=& \angle AEB    \end{array}

Therefore, \triangle ABE is an isosceles triangle.

Thus, AB=AE

(proved)

Congruent Triangles Maths Tuition: Solution

congruent-maths-tuition

Solution:

BS=BE+ES=ST+ES=ET

\angle DES=\angle ESA=90^\circ

BA=DT (given)

Thus, \triangle ASB is congruent to \triangle DET (RHS)

Hence \angle DTE=\angle SBA

Thus DT//BA (alt. angles)

(Proved)

By Pythagoras’ Theorem, we have

\begin{array}{rcl}DB&=&\sqrt{DE^2+BE^2}\\    &=&\sqrt{SA^2+ST^2}\\    &=&TA    \end{array}

Hence \triangle DEB and \triangle AST are congruent (SSS).

Hence \angle DBE=\angle STA

Thus DB//TA (alt. angles)

Therefore, ABDT is a parallelogram since it has two pairs of parallel sides.

(shown)