please help me to do this question
Answers
Answer:
Given: In ΔABC, ∠C = 90°
D, E, F are three points on BC such that they divided it into equal parts.
To prove: 8(AF^2+AD^2)=11AC^2+5AB^28(AF
2
+AD
2
)=11AC
2
+5AB
2
Proof:
D, E, F are three points on BC such that they divided it into equal parts. Therefore, BD = DE = EF = FC
DC=\dfrac{3}{4}BCDC=
4
3
BC
FC=\dfrac{1}{4}BCFC=
4
1
BC
In ΔAFC, ∠C = 90° , Using Pythagoras theorem
AF^2=AC^2+FC^2AF
2
=AC
2
+FC
2
AF^2=AC^2+\dfrac{BC^2}{16}AF
2
=AC
2
+
16
BC
2
---- (1) \because FC=\dfrac{1}{4}BC∵FC=
4
1
BC
In ΔADC, ∠C = 90° , Using Pythagoras theorem
AD^2=AC^2+DC^2AD
2
=AC
2
+DC
2
AD^2=AC^2+\dfrac{9BC^2}{16}AD
2
=AC
2
+
16
9BC
2
-----(2) \because FC=\dfrac{3}{4}BC∵FC=
4
3
BC
Add eq(1) and eq(2)
AF^2+AD^2=AC^2+\dfrac{BC^2}{16}+AC^2+\dfrac{9BC^2}{16}AF
2
+AD
2
=AC
2
+
16
BC
2
+AC
2
+
16
9BC
2
AF^2+AD^2=2AC^2+\dfrac{10BC^2}{16}AF
2
+AD
2
=2AC
2
+
16
10BC
2
AF^2+AD^2=\dfrac{32AC^2+10BC^2}{16}AF
2
+AD
2
=
16
32AC
2
+10BC
2
AF^2+AD^2=\dfrac{32AC^2+10AB^2-10AC^2}{16}AF
2
+AD
2
=
16
32AC
2
+10AB
2
−10AC
2
\because AB^2=AC^2+BC^2∵AB
2
=AC
2
+BC
2
AF^2+AD^2=\dfrac{22AC^2+10AB^2}{16}AF
2
+AD
2
=
16
22AC
2
+10AB
2
8(AF^2+AD^2)=11AC^2+5AB^28(AF
2
+AD
2
)=11AC
2
+5AB
2
hence proved