A square frame $ABCD$ contains a joint $P$. The cable $AP$ has the same length as side $AD$.

Question

A square frame $ABCD$ contains a joint $P$. The cable $AP$ has the same length as side $AD$. Segments $DP$ and $BP$ are also drawn. Diagonal $AC$ intersects $BP$ at $E$. It is given that $$\angle ADP=76^\circ.$$ Find $\angle PAB$. [3] Explain why triangle $ABP$ is isosceles. Hence find $\angle ABP$. [3] Find $\angle CEB$. [2] Give reasons for your answers.

Accepted Answer

(a) $62^\circ$ (b) $AB=AP$, so triangle $ABP$ is isosceles; $\angle ABP=59^\circ$. (c) $104^\circ$

Polygons - Properties of Special Quadrilaterals

Polygons - Properties of Special Quadrilaterals

Solution:

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