Diagonals AC and BD of quadrilateral ABCD intersect each other at P. Show that:
ar(APB)*ar(CPD) = ar(APD)*ar(BPC)
'ar' stands for 'area of'
Remember, ABCD is only an ordinary quadrilateral. It is not a trapezoid, parallelogram or other special quadrilaterals.
Hint: Draw AE and CF perpendicular to BD
I couldn't figure it out even with the hint.
2006-12-28 03:26:03 · 6 answers · asked by Akilesh - Internet Undertaker 7 in Science & Mathematics ➔ Mathematics
use the hint and draw the perpendiculars and using the formula area=(1/2)base*height
ar(APB)=(1/2)BP*AE
ar((CPD)=(1/2)DP*CF
ar(APD)=(1/2)DP*AE
ar((BPC)=(1/2)BP*CF
ar(APB)*ar(CPD)=(1/4)*BP*DP*AE*CF
ar(APD)*ar(BPC)=(1/4)*BP*DP*AE*CF
hence proved
2006-12-28 03:56:13 · answer #1 · answered by raj 7 · 1⤊ 1⤋
Let AP = x, PC = y, and s = altitude of ADC, and t = altitude of ABC. Then
ar(APB) = 1/2 x t
ar(BPC) = 1/2 y t
ar(APD) = 1/2 x s
ar(CPD) = 1/2 y t
We have: (1/2 x t )(1/2 y s) = (1/2 x s)(1/2 y t), which is
obviously true.
2006-12-28 03:57:19 · answer #2 · answered by Scythian1950 7 · 0⤊ 0⤋
AREA OF TRIANGLE = 1/2 x BASE x HEIGHT
ar(APB)*ar(CPD) = 1/2 x BP x AE x 1/2 x DP x CF
ar(APD)*ar(BPC) = 1/2 x DP x AE x 1/2 x BP x CF
REARRANGING, LEFT SIDE = RIGHT SIDE
2006-12-28 03:51:10 · answer #3 · answered by Sheen 4 · 0⤊ 0⤋
Area of traingle = 1/2 * base * height. We use AE and CF for altitudes.
Area of APB = a1 = 1/2 * BP * AE.
Area of CPD = a2 = 1/2 * PD * CF
Area of APD = a3 = 1/2 * PD * AE
Area of BPC = a4 = 1/2 * BP * CF
If we multiply, you see the result.
a1 * a2 = a3 * a4 = 1/4 * BP * PD * AE * CF.
2006-12-28 03:48:48 · answer #4 · answered by mlpkr 2 · 0⤊ 0⤋
since it is very difficult to explain things with out diagrams here kindly take help from your teacher.
2006-12-28 04:14:19 · answer #5 · answered by ram kumar 2 · 0⤊ 0⤋
Pythagoras theorem nothing more than that
2006-12-28 05:31:07 · answer #6 · answered by renganathan g 2 · 0⤊ 0⤋