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ΔABD and ΔPQD are similar, as the corresponding sides are parallel.
By BPT,
x / z = BD / QD => 1/x = QD /(z * BD)
ΔCDB and ΔPQB are similar, as corresponding sides are parallel.
By BPT,
y / z = BD / BQ => 1/y = BQ / (z * BD)
Add the two equations:
1/x + 1/y = (BQ + QD) / (z * BD) = BD / (z * BD)
= 1 / z
Hence proved
☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
By BPT,
x / z = BD / QD => 1/x = QD /(z * BD)
ΔCDB and ΔPQB are similar, as corresponding sides are parallel.
By BPT,
y / z = BD / BQ => 1/y = BQ / (z * BD)
Add the two equations:
1/x + 1/y = (BQ + QD) / (z * BD) = BD / (z * BD)
= 1 / z
Hence proved
☺ ☺ ☺ Hope this Helps ☺ ☺ ☺
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