Question

The following figure shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$ and $AD=BC$.

Prove that :

$i)\angle DAB=\angle CBA$

$ii)\angle ADC=\angle BCD$

$iii)AC=BD$

$iv)OA=OB$ and $OC=OD$

Answer 1

$\left(i\right)$ $\angle DAB=\angle CBA$

Construction :- Produce $AB$ to $E$ and draw $CE\parallel DA,MO\parallel AB$

Proof: So $AECD$ is a parallelogram.

In $△BEC$

$BC=CE$ ($AD=BC=CE$,opposite sides of parallelogram)

$\angle CBE=\angle CEB$ (opposite to equal sides) ---- $\left(1\right)$

$\angle ADC=\angle BEC$ (opposite angle of a parallelogram) ----- $\left(2\right)$

$\angle DAB+\angle ADC={180}^{\circ }$ (sum of co-interior angles) ------ $\left(3\right)$

$\angle ABC+\angle EBC={180}^{\circ }$ (L.P.P.)

$\angle ABC+\angle ADC={180}^{\circ }$ ---- $\left(4\right)$ from $\left(2\right)$

From equation $\left(3\right)$ and $\left(4\right)$

$\angle DAB=\angle CBA$

$\left(ii\right)$ $\angle BCD=\angle EBC$ ---- $\left(5\right)$ (alternate interior angles)

From equations $\left(1\right),\left(2\right),$ and $\left(5\right),$

$\angle ADC=\angle BCD$ proved

$\left(iii\right)$ In $△ABD$ and $\Delta BAC$

$\angle BAD=\angle ABC$

$AB=AB$ (common)

$AD=BC$ (Given)

$△ABD\cong △BAC$ (by SAS rule)

$BD=AC$ (By C.P.C.T)

$\left(iv\right)$ In $△ABD$

$MO\parallel AB$

$\frac{AM}{MD}=\frac{BO}{OD}$ --- $A$, (by lemma of B.P.T)

In $△ADC$

$MO\parallel DC$

$\frac{AM}{MD}=\frac{AO}{OC}$ ---- $B$ (by B.P.T.)

From equations $A$ and $B,$

$\frac{BO}{OD}=\frac{AO}{OC}$ ---- $C$

$\frac{BO}{OD}+1=\frac{AO}{OC}+1$

$\frac{BO+DO}{OD}=\frac{AO+OC}{OC}$

$\frac{BD}{OD}=\frac{AC}{OC}$

$OD=OC$ ($AC=BD$)

From equation $C,$

$OB=OA$ $(OD=OC)$