A triangle ABC is right angled at A . AL is drawn perpendicular to BC . Prove that angle BAL = ACB .
Answers
Answered by
12
HERE IS YOUR ANSWER____
*******____________******
Given: ABC is right triangle right angled at A and AL ⊥ BC
Now, in ΔBAL and ΔBCA
∠ABL = ∠CBA (Common)
∠BLA = ∠BAC = 90°
So, ΔBAL ∼ ΔBCA (by AA similarity criteria)
⇒ ∠BAL = ∠ACB (C.P.C.T)
@HOPE IT HELPS YOU!!
*******____________******
Given: ABC is right triangle right angled at A and AL ⊥ BC
Now, in ΔBAL and ΔBCA
∠ABL = ∠CBA (Common)
∠BLA = ∠BAC = 90°
So, ΔBAL ∼ ΔBCA (by AA similarity criteria)
⇒ ∠BAL = ∠ACB (C.P.C.T)
@HOPE IT HELPS YOU!!
rpreddy62p5s5jg:
there is no aa criteria
Answered by
7
No need to do any congruency stuff.
Also It is not given that BL = CL. It can be simply done,
Angle BAL = BAC - LAC
=90° - LAC (1)
In triangle ADC, by angle sum property,
ALC + LAC + ACL = 180°
90° + LAC + ACB = 180° (ACL=ACB) ( common)
ACB = 90° - LAC (2)
From (1) and (2),
BAL = ACB = 90° - LAC
Hence Proved
Similar questions