ABC is a right angled triangle, right angle at A. If AD is perpendicular to BC, then
BC . CD = BC . BC
AB . AC = BC . BC
BD . CD = AD . AD
AB . AC = AD . AD
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Answer:
AB , AC = AD, AD
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Question-
In the following figure, AB = BC and AD = CD. Show that BD bisects AC at right angles.
Answer-
We are required to prove ∠BEA = ∠BEC = 90° and AE = EC.Consider ∆ABD and ∆CBD,
AB = BC (Given)
AD = CD (Given)
BD = BD (Common)
Therefore, ∆ABD ≅ ∆CBD (By SSS congruency)
∠ABD = ∠CBD (CPCT)
Now, consider ∆ABE and ∆CBE,
AB = BC (Given)
∠ABD = ∠CBD (Proved above)
BE = BE (Common)
Therefore, ∆ABE≅ ∆CBE (By SAS congruency)
∠BEA = ∠BEC (CPCT)
And ∠BEA +∠BEC = 180° (Linear pair)
2∠BEA = 180° (∠BEA = ∠BEC)
∠BEA = 180°/2 = 90° = ∠BEC
AE = EC (CPCT)
Hence, BD is a perpendicular bisector of AC.
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