Question

In the given figure, point X is the midpoint of side BC , seg XZ II seg AB and seg YZ II seg AX. Prove that YC =1/4 BC ​

Answer 1

$YC=\frac{1}{4}BC$

Step-by-step explanation:

From the figure,

AB ║ ZX and AC is a transverse.

Therefore, ∠BAC ≅ ∠XZC [Corresponding angles]

∠ACB is common in both the triangles.

Therefore, ΔABC and ΔZXC are similar [AA property of similarity]

By the property of similarity,

$\frac{BC}{XC}=\frac{AC}{ZC}$ -----(1)

Similarly in ΔXAC and ΔYZC,

It is given that AX║ZY and AC is a transverse.

Therefore, ∠XAC ≅ ∠YZC [Corresponding angles]

∠ACB is common in both the triangles.

By the property of AA of similarity ΔXAC and ΔYZC will be similar.

Now by the property of similarity corresponding sides will be in the same ratio.

$\frac{XC}{YC}=\frac{AC}{ZC}$ ------(2)

From equation (1) and (2),

$\frac{BC}{XC}=\frac{XC}{YC}$

It is given that 2XC = BC [Given that X is the midpoint of BC]

$\frac{BC}{\frac{BC}{2}}=\frac{\frac{BC}{2}}{YC}$

$\frac{BC}{2YC}=2$

$YC=\frac{1}{4}BC$

Hence proved.

Learn more about similar triangles from https://brainly.in/question/1296566

Answer 2

Answer:

perfect answer. ...........................