Question

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.

Answer 1

Given: In two angles $\angle ABCand\angle DEFAB\perp DEandBC\perp EF$
To prove: $\angle ABC+\angle DEF={180}^{0}or\angle ABC=\angle DEF$
Construction: Produce the sides DE and EF of $\angle$ DEF, to meet the sides of $\angle$ ABC at H and G.

Proof: In figure (i) BGEH is a quadrilateral $\angle BHE={90}^{0}and\angle BGE={90}^0$
But sum of angles of a quadrilateral is ${360}^0$
$\therefore \angle HBG+\angle HEG={360}^0-({90}^0+{90}^0)$
=${360}^0-{180}^0={180}^0$
$\therefore \angle ABCand\angle DEF$ are supplementary In figure (ii) in quadrilateral BGEH,
$\angle BHE={90}^{0}and\angle HEG={90}^0$
$\therefore \angle HBG+\angle HEG={360}^0-({90}^0+{90}^0)$
= ${360}^0-{180}^0={180}^0$ ...(i)
But $\angle HEF+\angle HEG={180}^0$ ...(ii)
(Linear pair)
From (i) and (ii)
$\therefore \angle HEF=\angle HBG$
$\Rightarrow \angle DEF=\angle ABC$
Hence $\angle ABCand\angle DEF$ are equal or supplementary