Question

$P$ is the mid-point of the hypotenuse $AB$ of the right -angled triangle $ABC.$ then, $:AB=2CP.$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given: $△ABC$, P is mid point of AB, $\angle C={90}^{\circ }$
To prove: $PA=PB=\frac{1}{2}AB$
Construction: Draw $PK\parallel BC$
Since, $PK\parallel BC$ with transversal AC,
$\angle 1=\angle C={90}^{\circ }$
Also, $\angle 1+\angle 2=180$
$\angle 2=180-\angle 1=180-90={90}^{\circ }$
Now, In $△APK$ and $△CPK$,
$KP=KP$ (Common)
$\angle 1=\angle 2={90}^{\circ }$
$AK=KC$ (Since, $KP\parallel BC$ and P is mid point of AB)
$△APK\cong △CPK$ (SAS rule)
Therefore, $PA=PC$
$PA=\frac{1}{2}AB$
$PA=PC=\frac{1}{2}AB$
$AB=2PC$