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Step-by-step explanation:
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AB, CD, PQ are perpendicular to BD. AB = x, CD = y amd PQ = Z prove that \frac{1}{x} + \frac{1}{y} = \frac{1}{z}
x
1
+
y
1
=
z
1
430 viewed last edited 4 months ago
Mathematicssimilar trianglesUsing similar triangles propertiesProve that the triangle are similar using AAA theoremratios of corresponding sides are equalhigh-school
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Krishna(Expert, Educator @Qalaxia)(last edited 4 months ago)
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Step 1: Explore the given figure and note down the given measurement
NOTE: AB, CD, PQ are perpendicular to BD
AB = x, CD = y and PQ = z
Step 2: Prove that the ∆ABD and ∆PQD are similar
EXPLANATION: \angle ∠ ABD = \angle ∠ PQD = 90°
\angle ∠ADB = \angle ∠PDQ ( common angle )
By AAA similarity ,
We can conclude that ∆ABD ~ ∆PQD
Step 3: Prove that ∆CDB and ∆PQB are similar
EXPLANATION: \angle ∠ CDB = \angle ∠ PQB = 90°
\angle ∠ CBD = \angle ∠ PBQ ( common angle )
We can conclude that ∆CDB ~ ∆PQB ( A.A similarity )
Step 4: Write the ratios of the lengths of similar triangles corresponding sides.
NOTE: The ratios of the lengths of similar triangles corresponding sides
are equal.
From ∆ABD ~ ∆PQD
We can write \frac{PQ}{AB} = \frac{QD}{BD}
AB
PQ
=
BD
QD
\frac{z}{x} = \frac{QD}{BD}
x
z
=
BD
QD
.....................(1)
From ∆CDB ~ ∆PQB
we can write \frac{PQ}{CD} = \frac{BQ}{BD}
CD
PQ
=
BD
BQ
\frac{z}{y} = \frac{BQ}{BD}
y
z
=
BD
BQ
........................(2)
Step 5: Add equation (1) and (2)
From ( 1 ) and ( 2 ) we get
\frac{z}{x} + \frac{z}{y} = \frac{QD}{BD} + \frac{BQ}{BD}
x
z
+
y
z
=
BD
QD
+
BD
BQ
z(\frac{1}{x} + \frac{1}{y}) = \frac{QD + BQ}{BD} z(
x
1
+
y
1
)=
BD
QD+BQ
z(\frac{1}{x} + \frac{1}{y}) = \frac{BD}{BD} z(
x
1
+
y
1
)=
BD
BD
(Since BD = QD + BQ)
z(\frac{1}{x} + \frac{1}{y}) = 1 z(
x
1
+
y
1
)=1
\frac{1}{x} + \frac{1}{y} = \frac{1}{z}
x
1
+
y
1
=
z
1
Hence proved