Question

If BC is the greatest side of ΔABC , and D,E are points on BC,CA respectively. Prove that BC ≥ EF.

Answer 1

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.Lemma: Let PQR be a triangle, and let X be in the interval PR, and not at an endpoint. Then QX<QP or QX<QR (or both). More compactly, QX<max(QP,QR).

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.Lemma: Let PQR be a triangle, and let X be in the interval PR, and not at an endpoint. Then QX<QP or QX<QR (or both). More compactly, QX<max(QP,QR).Proof: Since angles QXP and QXR add up to 180∘, one at least is ≥90∘. So ∠QXP is the largest angle in △QXP, or ∠QXR is the largest angle in △QXR, or both. If ∠QXP is the largest angle in △QXP, then QP>QX. Similarly, if ∠QXR is the largest angle in △QXR, then QR>QX.

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.Lemma: Let PQR be a triangle, and let X be in the interval PR, and not at an endpoint. Then QX<QP or QX<QR (or both). More compactly, QX<max(QP,QR).Proof: Since angles QXP and QXR add up to 180∘, one at least is ≥90∘. So ∠QXP is the largest angle in △QXP, or ∠QXR is the largest angle in △QXR, or both. If ∠QXP is the largest angle in △QXP, then QP>QX. Similarly, if ∠QXR is the largest angle in △QXR, then QR>QX.Now we look at our triangle ABC. Suppose that D=B. If E=C, then DE=BE=BC. If E≠C, then by the Lemma BE (that is, DE) is <max(BC,BA)=BC.

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.Lemma: Let PQR be a triangle, and let X be in the interval PR, and not at an endpoint. Then QX<QP or QX<QR (or both). More compactly, QX<max(QP,QR).Proof: Since angles QXP and QXR add up to 180∘, one at least is ≥90∘. So ∠QXP is the largest angle in △QXP, or ∠QXR is the largest angle in △QXR, or both. If ∠QXP is the largest angle in △QXP, then QP>QX. Similarly, if ∠QXR is the largest angle in △QXR, then QR>QX.Now we look at our triangle ABC. Suppose that D=B. If E=C, then DE=BE=BC. If E≠C, then by the Lemma BE (that is, DE) is <max(BC,BA)=BC.Suppose now that D≠B. The case E=C is easy.

We assume as known that in any triangle PQR, the largest side is opposite the largest angle, and conversely.Lemma: Let PQR be a triangle, and let X be in the interval PR, and not at an endpoint. Then QX<QP or QX<QR (or both). More compactly, QX<max(QP,QR).Proof: Since angles QXP and QXR add up to 180∘, one at least is ≥90∘. So ∠QXP is the largest angle in △QXP, or ∠QXR is the largest angle in △QXR, or both. If ∠QXP is the largest angle in △QXP, then QP>QX. Similarly, if ∠QXR is the largest angle in △QXR, then QR>QX.Now we look at our triangle ABC. Suppose that D=B. If E=C, then DE=BE=BC. If E≠C, then by the Lemma BE (that is, DE) is <max(BC,BA)=BC.Suppose now that D≠B. The case E=C is easy.If B≠D and E≠C, draw the line BE. By the Lemma, we have BE<BC. Again by the Lemma, DE<max(BE,CE)<BC. This completes the proof.

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