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Preeti Dabral 6 days, 1 hour ago

Given: ABCD is a square, where {tex}\angle{/tex}PQR = 90^{o }and PB = QC = DR

To prove: (i) QB = RC (ii) PQ = QR

(iii) {tex}\angle{/tex}QPR = 45^{o}

Proof:

- Here,

BC = CD … (Sides of square)

CQ = DR … (Given)

BC = BQ + CQ

{tex}\therefore{/tex} CQ = BC − BQ

{tex}\therefore{/tex} DR = BC – BQ .......(1)

Also,

CD = RC + DR

{tex}\therefore{/tex} DR = CD - RC = BC - RC ........(2)

From (1) and (2), we have,

BC - BQ = BC - RC

{tex}\therefore{/tex} BQ = RC - Now in {tex}\triangle{/tex}RCQ and {tex}\triangle{/tex}QBP

we have, PB = QC … (Given)

BQ = RC … [from (i)]

{tex}\angle{/tex}RCQ = {tex}\angle{/tex}QBP … 90^{o}each

Hence by SAS(Side-Angle-Side) congruence rule,

{tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP

{tex}\therefore{/tex} QR = PQ … (by cpct) - {tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP and QR = PQ … [from (2)]

{tex}\therefore{/tex} In {tex}\triangle{/tex}RPQ,

{tex}\angle{/tex}QPR = {tex}\angle{/tex}QRP ={tex}\frac{1}{2}{/tex}(180^{o}- 90^{o}) = {tex}\frac{90}{2}{/tex}= 45^{o}

{tex}\therefore{/tex} {tex}\angle{/tex}QPR = 45^{o }

Posted by Hena Kausar 6 days, 17 hours ago

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Posted by Jinali Zaveri 1 week ago

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Preeti Dabral 1 week ago

Given: In right triangle ABC, right angled at C. M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.

To Prove:

- {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD
- {tex}\angle{/tex}DBC is a right angle
- {tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB
- CM = {tex}\frac {1} {2}{/tex}AB

Proof:

- In {tex}\triangle{/tex}AMC and {tex}\triangle{/tex}BMD

AM = BM ...[As M is the mid-point]

CM = DM ...[Given]

{tex}\angle{/tex}AMC = {tex}\angle{/tex}BMD ...[Vertically opposite angles]

{tex}\therefore{/tex} {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD proved ...[SAS property] ...(1) - {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]

{tex}\angle{/tex}ACM = {tex}\angle{/tex}BDM ...[c.p.c.t.]

These are alternate interior angles and they are equal.

{tex}\therefore{/tex} AC {tex}\|{/tex} BD

As AC {tex}\|{/tex} BD and transversal BC intersects them

{tex}\therefore{/tex} {tex}\angle{/tex}DBC + {tex}\angle{/tex}ACB = 180° ...[Sum of the consecutive interior angles of the transversal]

{tex}\angle{/tex}DBC + 90° = 180°

{tex}\angle{/tex}DBC = 180° - 90° = 90°

{tex}\angle{/tex}DBC is a right angle proved. - {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]

{tex}\therefore{/tex} AC = BD ...[c.p.c.t.] ...(2)

In DDBC and DACB

BC = CB ...[Common]

{tex}\angle{/tex}DBC = {tex}\angle{/tex}ACB ...[each = 90°^{ }as proved above]

BD = CA ...[From (2)]

{tex}\therefore{/tex} {tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB ...[SAS property] - DDBC {tex}\cong{/tex} DACB ...[As proved in (iii)]

{tex}\therefore{/tex} DC = AB ...[c.p.c.t.]

{tex}\therefore{/tex} 2CM = AB ...[DM = CM = {tex}\frac {1} {2}{/tex} DC]

{tex}\therefore{/tex} CM ={tex}\frac {1} {2}{/tex} AB

Posted by Sonali Verma 6 days, 23 hours ago

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Preeti Dabral 6 days, 23 hours ago

Let a square ABCD in which L, M, N & O are the midpoints.

In {tex}\triangle{/tex}AML and {tex}\triangle{/tex}CNO

AM = CO (AB = DC and M and O are the midpoints)

AL = CN (AD = BC and L and N are the midpoints)

{tex}\angle{/tex} MAL = {tex}\angle{/tex}NCO (all angles of a square = 90°)

by AAS criteria

{tex}\triangle{/tex}AML {tex}\cong{/tex} {tex}\triangle{/tex}CNO

{tex}\therefore{/tex} ML = ON (CPCT)

similarly {tex}\triangle{/tex}MBN {tex}\cong{/tex} {tex}\triangle{/tex}LDO

now,

in {tex}\triangle{/tex}AML,

{tex}\angle{/tex}AML = {tex}\angle{/tex}ALM (AM = AL)

= 45°

similarly in {tex}\triangle{/tex}LDO

{tex}\angle{/tex}DLO = 45°

{tex}\therefore{/tex} {tex}\angle{/tex} MLO = 90°

by the properties of SQUARE

All sides are equal and angles are 90°

Posted by Harshdeep Raj 1 week ago

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Preeti Dabral 6 days, 22 hours ago

Factor Theorem is generally applied to factoring and finding the roots of polynomial equations. It is the reverse form of the remainder theorem. Problems are solved based on the application of synthetic division and then to check for a zero remainder.

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