In the figure point X is the midpoint of side BC seg XZ is drawn parallel to side AB seg YZ is parallel to seg AX PROVE: YC=1\4 BC
Answers
Answer:
Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.
Given:
BC = 9, AB = 5, PQ = 6 and QR = 10.
∴
A(△ABC)
A(△PQR)
=
AB×BC
PQ×QR
=
5×9
6×10
=
3
4
Step-by-step explanation:
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Answer:
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}
XC
BC
=
ZC
AC
-----(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}
YC
XC
=
ZC
AC
------(2)
From equation (1) and (2),
\frac{BC}{XC}=\frac{XC}{YC}
XC
BC
=
YC
XC
It is given that 2XC = BC [Given that X is the midpoint of BC]
\frac{BC}{\frac{BC}{2}}=\frac{\frac{BC}{2}}{YC}
2
BC
BC
=
YC
2
BC
\frac{BC}{2YC}=2
2YC
BC
=2
YC=\frac{1}{4}BCYC=
4
1
BC