Question

In the triangle ABC, ADIBC. Show that ABP – BD2 = AC? – DC2
20. K is a point inside the rectangle ARC Show that KA2KC2 - YP2
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Answer 1

Explanation:

Given: AB = AC, D is the point in the interior of ∆ABC such that ∠DBC = ∠DCB.

To Prove: AD bisects ∠BAC of ∆ABC.

Proof: In ∆DBC,

∵ ∠DBC = ∠DCB ...(1) | Given

∴ DB = DC ...(2)

| Sides opposite to equal angles of a triangle are equal

In ∆ABC,

∵ AB = AC | Given

∴ ∠ABC = ∠ACB

∴ ∠ABC = ∠ACB ...(3)

| Angles opposite to equal sides of a triangle are equal

Subtracting (1) from (3), we get,

∠ABC - ∠DBC = ∠ACB - ∠DCB

⇒ ∠ABD = ∠ACD ...(4)

In ∆ADB and ∆ADC,

AB = AC | Given

DB = DC | Proved in (2)

∠ABD = ∠ACD | Proved in (4)

∴ ∆ADB ≅ ∆ADC

| SAS congruence rule

∴ ∠DAB = ∠DAC | CPCT

⇒ AD bisects ∠BAC of ∆ABC.