Question

In a triangle ABC, BD bisects ∟B and BD ꓕ AC. If the length of the sides of the triangle are AB = 3x + 1, BC = 5y – 2, AD = x + 1 and DC = y + 2, find the values of x and y.

Answer 1

Answer:

In △ABD and △CBD

BD=BD (common)

∠ADB=∠CDB (each 90

∘

)

∠ABD=∠CDB ( BD bisect ∠B)

△ABD≅△CDB (by ASA)

⇒3x+1=5y−2 (CPCT)

⇒x=

3

5y−3

.....(1)

⇒x+1=y+2 (CPCT)

⇒x=y+1 .........(2)

From (1) and (2)

⇒5y−3=3(y+1)

⇒5y−3y=3+3

⇒2y=6

⇒y=3

put y=3 in (2)

x=3+1

∴x=4

Step-by-step explanation:

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