In the given figure, XN is parallel to CA and XM is parallel to BA. T is a point on CB produced. Prove that TX 2 =TB.TC.
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41
XN II MX
NB II MX ( A - N - B)
On transversal TX
∠TBN Congruent to ∠TXM
In ΔTBN and ΔTXM
∠TBN congruent ∠TXM [From above]
∠BTN congruent ∠XTM
ΔTBN is similar to ΔTXM
TB/TX = TN/TM
NX II AC
NX II MC
On traversal TC,
∠TXN congruent to ∠TCM
∠TXN congruent ∠TCM
∠NTX congruent ∠MTC
ΔTXN is similar to ΔTCM
TX/TC = TN/ TM
TB/ TX = TX/TC
TX²= TB x TC
NB II MX ( A - N - B)
On transversal TX
∠TBN Congruent to ∠TXM
In ΔTBN and ΔTXM
∠TBN congruent ∠TXM [From above]
∠BTN congruent ∠XTM
ΔTBN is similar to ΔTXM
TB/TX = TN/TM
NX II AC
NX II MC
On traversal TC,
∠TXN congruent to ∠TCM
∠TXN congruent ∠TCM
∠NTX congruent ∠MTC
ΔTXN is similar to ΔTCM
TX/TC = TN/ TM
TB/ TX = TX/TC
TX²= TB x TC
Answered by
2
Answer:
tx2=tb*tc
Step-by-step explanation:
XN II MX
NB II MX ( A - N - B)
On transversal TX
∠TBN Congruent to ∠TXM
In ΔTBN and ΔTXM
∠TBN congruent ∠TXM [From above]
∠BTN congruent ∠XTM
ΔTBN is similar to ΔTXM
TB/TX = TN/TM
NX II AC
NX II MC
On traversal TC,
∠TXN congruent to ∠TCM
∠TXN congruent ∠TCM
∠NTX congruent ∠MTC
ΔTXN is similar to ΔTCM
TX/TC = TN/ TM
TB/ TX = TX/TC
TX²= TB x TC
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