Congruent Triangles Maths Tuition: Solution




\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)


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.


Author: mathtuition88

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