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Given DE//QR and
AP and BP are bisectors of angle(EAB) and angle(RBA) respectively
DE//QR, n be the transversal
=>EAB+RBA=180° (Co-interior angles)
(EAB+RBA)/2=180°/2
EAB/2+RBA/2=90°
PAB+PBA=90° (GIVEN) -------(1)
In ∆PAB
APB+PAB+ABP=180°
APB+90°=180° ( From (1) )
APB=180°-90°
APB=90°
HOPE IT HELPS
AP and BP are bisectors of angle(EAB) and angle(RBA) respectively
DE//QR, n be the transversal
=>EAB+RBA=180° (Co-interior angles)
(EAB+RBA)/2=180°/2
EAB/2+RBA/2=90°
PAB+PBA=90° (GIVEN) -------(1)
In ∆PAB
APB+PAB+ABP=180°
APB+90°=180° ( From (1) )
APB=180°-90°
APB=90°
HOPE IT HELPS
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